`(13pi)/12` Find the exact values of the sine, cosine, and tangent of the angle.

mathace | Certified Educator

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`(13pi)/12=(3pi)/4+pi/3`

`sin(u+v)=sin(u)cos(v)+cos(u)sin(v)`

`sin((3pi)/4+pi/3)=sin((3pi)/4)cos(pi/3)+cos(pi/3)sin((3pi)/4)`

`sin((3pi)/4+pi/3)=(sqrt2/2)(1/2)+(-sqrt2/2)(sqrt3/2)=sqrt2/4(1-sqrt3)`

`cos(u+v)=cos(u)cos(v)-sin(u)sin(v)`

`cos((3pi)/4+pi/3)=cos((3pi)/4)cos(pi/3)-sin((3pi)/4)sin(pi/3)`

`cos((3pi)/4+pi/3)=(-sqrt2/2)(1/2)-(sqrt2/2)(sqrt3/2)=-sqrt2/4(1+sqrt3)`

`tan(u+v)=(tan(u)+tan(v))/(1-tan(u)tan(v))`

`tan((3pi)/4+pi/3)=(tan((3pi)/4)+tan(pi/3))/(1-tan((3pi)/4)tan(pi/3))=((-1)+sqrt3)/(1-(-1)(sqrt3))=(-1+sqrt3)/(1+sqrt3)`

Rationalize the denominator.

`(-1+sqrt3)/(1+sqrt3)*(1-sqrt3)/(1-sqrt3)=(-1+2sqrt3-3)/(1-3)=(-4+2sqrt3)/(-2)=2-sqrt3`

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gsarora17 | Certified Educator

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starTop subjects are Math, Science, and Business

`sin((13pi)/12)=sin(pi/2+pi/3+pi/4)`

As we know that `sin(pi/2+theta)=cos(theta)`

`:.sin((13pi)/12)=cos(pi/3+pi/4)`

Now use the identity `cos(x+y)=cos(x)cos(y)-sin(x)sin(y)`

`sin((13pi)/12)=cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)`

`sin((13pi)/12)=((1/2)(1/sqrt(2))-((sqrt(3)/2)(1/sqrt(2)))`

`sin((13pi)/12)=(1-sqrt(3))/(2sqrt(2))`

`sin((13pi)/12)=(sqrt(2)-sqrt(6))/4`

`cos((13pi)/12)=cos(pi/2+pi/3+pi/4)`

We know that `cos(pi/2+theta)=-sin(theta)`

`:.cos((13pi)/12)=-sin(pi/3+pi/4)`

using identity `sin(x+y)=sin(x)cos(y)+cos(x)sin(y)`

`cos((13pi)/12)=-(sin(pi/3)cos(pi/4)+cos(pi/3)sin(pi/4))`

`=-(sqrt(3)/2*1/sqrt(2)+1/2*1/sqrt(2))`

`=-(sqrt(3)+1)/(2sqrt(2))`

`=-(sqrt(6)+sqrt(2))/4`

`tan((13pi)/12)=sin((13pi)/12)/cos((13pi)/12)`

plug in the values evaluated above,

`tan((13pi)/12)=((sqrt(2)-sqrt(6))/4)/(-(sqrt(6)+sqrt(2))/4)`

`=(sqrt(2)-sqrt(6))/(-(sqrt(6)+sqrt(2)))`

rationalizing the denominator,

`=((sqrt(2)-sqrt(6))(sqrt(6)-sqrt(2)))/(-(6-2))`

`=(sqrt(12)-2-6+sqrt(12))/(-4)`

`=(2sqrt(12)-8)/(-4)`

`=(2*2sqrt(3)-8)/(-4)`

`=2-sqrt(3)`

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